[LeetCode] Interval Cancellation
![[LeetCode] Interval Cancellation](/assets/img/algorithm.png)
PROBLEMS
Description
Given a function fn, an array of arguments args, and an interval time t, return a cancel function cancelFn.
After a delay of cancelTimeMs, the returned cancel function cancelFn will be invoked.
setTimeout(cancelFn, cancelTimeMs)
The function fn should be called with args immediately and then called again every t milliseconds until cancelFn is called at cancelTimeMs ms.
Example
Example 1:
Input: fn = (x) => x _ 2, args = [4], t = 35
Output:
[{“time”: 0, “returned”: 8}, {“time”: 35, “returned”: 8}, {“time”: 70, “returned”: 8}, { “time”: 105, “returned”: 8}, {“time”: 140, “returned”: 8}, {“time”: 175, “returned”: 8} ]
Explanation:
const cancelTimeMs = 190;
const cancelFn = cancellable((x) => x _ 2, [4], 35);
setTimeout(cancelFn, cancelTimeMs);
Every 35ms, fn(4) is called. Until t=190ms, then it is cancelled.
1st fn call is at 0ms. fn(4) returns 8.
2nd fn call is at 35ms. fn(4) returns 8.
3rd fn call is at 70ms. fn(4) returns 8.
4th fn call is at 105ms. fn(4) returns 8.
5th fn call is at 140ms. fn(4) returns 8.
6th fn call is at 175ms. fn(4) returns 8.
Cancelled at 190ms
Example 2:
Input: fn = (x1, x2) => (x1 _ x2), args = [2, 5], t = 30
Output:
[{“time”: 0, “returned”: 10}, {“time”: 30, “returned”: 10}, {“time”: 60, “returned”: 10}, {“time”: 90, “returned”: 10}, {“time”: 120, “returned”: 10}, {“time”: 150, “returned”: 10} ]
Explanation:
const cancelTimeMs = 165;
const cancelFn = cancellable((x1, x2) => (x1 _ x2), [2, 5], 30)
setTimeout(cancelFn, cancelTimeMs)
Every 30ms, fn(2, 5) is called. Until t=165ms, then it is cancelled.
1st fn call is at 0ms
2nd fn call is at 30ms
3rd fn call is at 60ms
4th fn call is at 90ms
5th fn call is at 120ms
6th fn call is at 150ms
Cancelled at 165ms
Example 3:
Input: fn = (x1, x2, x3) => (x1 + x2 + x3), args = [5, 1, 3], t = 50
Output:
[
{“time”: 0, “returned”: 9},
{“time”: 50, “returned”: 9},
{“time”: 100, “returned”: 9},
{“time”: 150, “returned”: 9}
]
Explanation:
const cancelTimeMs = 180;
const cancelFn = cancellable((x1, x2, x3) => (x1 + x2 + x3), [5, 1, 3], 50)
setTimeout(cancelFn, cancelTimeMs)
Every 50ms, fn(5, 1, 3) is called. Until t=180ms, then it is cancelled.
1st fn call is at 0ms
2nd fn call is at 50ms
3rd fn call is at 100ms
4th fn call is at 150ms
Cancelled at 180ms
Constraints
- fn is a function
- args is a valid JSON array
- 1 <= args.length <= 10
- 30 <= t <= 100
- 10 <= cancelTimeMs <= 500
Solution
- Runtime: 63ms
- Memory: 49.82MB
/**
* @param {Function} fn
* @param {Array} args
* @param {number} t
* @return {Function}
*/
var cancellable = function(fn, args, t) {
fn(...args);
const timeout = setInterval(() => {
fn(...args)
}, t);
return function cancel() {
clearInterval(timeout);
}
};
/**
* const result = [];
*
* const fn = (x) => x * 2;
* const args = [4], t = 35, cancelTimeMs = 190;
*
* const start = performance.now();
*
* const log = (...argsArr) => {
* const diff = Math.floor(performance.now() - start);
* result.push({"time": diff, "returned": fn(...argsArr)});
* }
*
* const cancel = cancellable(log, args, t);
*
* setTimeout(cancel, cancelTimeMs);
*
* setTimeout(() => {
* console.log(result); // [
* // {"time":0,"returned":8},
* // {"time":35,"returned":8},
* // {"time":70,"returned":8},
* // {"time":105,"returned":8},
* // {"time":140,"returned":8},
* // {"time":175,"returned":8}
* // ]
* }, cancelTimeMs + t + 15)
*/
Retrospect
이전에 풀었던 setTimeout이랑 비슷한 유형의 문제같다는 생각을 했다. setTimeout으로 cancelTimeMs변수의 시간이 끝난 후, cancelFn를 실행시키면 되는 문제였기 때문에 cancelFn안에서는 interval을 멈추는 함수를 실행해주었다.
제일 처음 cancellable함수에 값을 넘겨주면 cancellable안쪽에서는 setInterval함수를 호출해서 계속해서 interval을 하게끔 만들어주었다. 처음에 cancellable함수에 생설될때 time은 0이 될텐데 setInterval로만 실행하게 되면 time이 0일때는 실행이 안되어서 fn(...args)로 따로 빼서 한번 실행해주었다.
마지막 반환값으로는 interval을 멈추는 clearInterval함수를 호출하여 외부에서 setTimeout이 호출될때 interval을 중지시키므로써 문제에서 요구하는 답을 추출하게끔 했다.
